Giải bài tập 4. 10 trang 18 SGK Toán 12 tập 2 - Kết nối tri thức — Không quảng cáo

Đề bài

Tính:

a) $\int\limits_0^3 {{{\left( {3x - 1} \right)}^2}dx}$;

b) $\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \sin x} \right)dx}$;

c) $\int\limits_0^1 {\left( {{e^{2x}} + 3{x^2}} \right)dx}$;

d) $\int\limits_{ - 1}^2 {\left| {2x + 1} \right|dx}$.

Lời giải chi tiết

a) $\int\limits_0^3 {{{\left( {3x - 1} \right)}^2}dx}  = \int\limits_0^3 {\left( {9{x^2} - 6x + 1} \right)dx}  = 9\int\limits_0^3 {{x^2}dx}  - 6\int\limits_0^3 {xdx}  + \int\limits_0^3 {dx}$

$= 3{x^3}\left| \begin{array}{l}3\\0\end{array} \right. - 3{x^2}\left| \begin{array}{l}3\\0\end{array} \right. + x\left| \begin{array}{l}3\\0\end{array} \right. = 81 - 27 + 3 = 57$

b) $\int\limits_0^{\frac{\pi }{2}} {\left( {1 + \sin x} \right)dx}  = \int\limits_0^{\frac{\pi }{2}} {dx}  + \int\limits_0^{\frac{\pi }{2}} {\sin xdx}  = x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. - \cos x\left| \begin{array}{l}\frac{\pi }{2}\\0\end{array} \right. = \frac{\pi }{2} + 1$

c) $\int\limits_0^1 {\left( {{e^{2x}} + 3{x^2}} \right)dx}  = \int\limits_0^1 {{{\left( {{e^2}} \right)}^x}dx}  + 3\int\limits_0^1 {{x^2}dx}  = \frac{{{e^{2x}}}}{{\ln {e^2}}}\left| \begin{array}{l}1\\0\end{array} \right. + {x^3}\left| \begin{array}{l}1\\0\end{array} \right. = \frac{{{e^2}}}{2} - \frac{1}{2} + 1 = \frac{{{e^2}}}{2} + \frac{1}{2}$

d) $\int\limits_{ - 1}^2 {\left| {2x + 1} \right|dx}  = \int\limits_{ - 1}^{\frac{{ - 1}}{2}} {\left| {2x + 1} \right|dx}  + \int\limits_{\frac{{ - 1}}{2}}^2 {\left| {2x + 1} \right|dx}  =  - \int\limits_{ - 1}^{\frac{{ - 1}}{2}} {\left( {2x + 1} \right)dx}  + \int\limits_{\frac{{ - 1}}{2}}^2 {\left( {2x + 1} \right)dx}$

$=  - \left( {{x^2} + x} \right)\left| \begin{array}{l}\frac{{ - 1}}{2}\\ - 1\end{array} \right. + \left( {{x^2} + x} \right)\left| \begin{array}{l}2\\\frac{{ - 1}}{2}\end{array} \right. =  - \left[ {{{\left( {\frac{{ - 1}}{2}} \right)}^2} - \frac{1}{2} - {{\left( { - 1} \right)}^2} + 1} \right] + \left[ {{2^2} + 2 - {{\left( {\frac{{ - 1}}{2}} \right)}^2} + \frac{1}{2}} \right]$

$= \frac{1}{4} + \frac{{25}}{4} = \frac{{13}}{2}$